The problem is to calculate the following limit: $$\lim_{n \to \infty} \sum_{k=n}^{5n} \binom{k-1}{n-1} (\frac{1}{5})^{n} (\frac{4}{5})^{k-n}$$ I tried to reduce the expression using the generating function and it became $$\lim_{n \to \infty} (\frac{1}{5})^{n}\Bigl(\sum_{k=n-1}^{\infty}\binom{k}{n-1}(\frac{4}{5})^k - \sum_{k=5n}^{\infty}\binom{k}{n-1}(\frac{4}{5})^k\Bigr)$$ And now, i can't calculate the last sum, maybe i was wrong from the start. Help me, please.
Thank You in advance.
Let $4/5=x$, then the given sum $S$ is $$S=(1-x)^n x^{-n} \sum_{k=n}^{5n}{k-1 \choose n-1} x^k= (1-x)^n x^{-n}[1+nx+n(n+1) \frac{x^2}{2!}+n(n+1)(n+2) \frac{x^3}{3!}+....ad ~inf] x^n = (1-x)^{n} x^{-n} (1-x)^{-n} x^n=1.$$
Hence the given limit is 1, as $n \rightarrow \infty$